Is ${53705}$ divisible by $3$ ?
A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {53705}= &&{5}\cdot10000+ \\&&{3}\cdot1000+ \\&&{7}\cdot100+ \\&&{0}\cdot10+ \\&&{5}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {53705}= &&{5}(9999+1)+ \\&&{3}(999+1)+ \\&&{7}(99+1)+ \\&&{0}(9+1)+ \\&&{5} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {53705}= &&\gray{5\cdot9999}+ \\&&\gray{3\cdot999}+ \\&&\gray{7\cdot99}+ \\&&\gray{0\cdot9}+ \\&& {5}+{3}+{7}+{0}+{5} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first four terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${53705}$ is divisible by $3$ if ${ 5}+{3}+{7}+{0}+{5}$ is divisible by $3$ Add the digits of ${53705}$ $ {5}+{3}+{7}+{0}+{5} = {20} $ If ${20}$ is divisible by $3$ , then ${53705}$ must also be divisible by $3$ ${20}$ is not divisible by $3$, therefore ${53705}$ must not be divisible by $3$.